Question

A plane electromagnetic wave travelling in a non-magnetic medium is given by

$$E = \left( {9 \times 10^8 \,NC^{ - 1} } \right)\sin \left[ {\left( {9 \times 10^8 \,rad\,s^{ - 1} } \right)t - \left( {6\,m^{ - 1} } \right)x} \right]$$

where x is in meters and t is in second. What will be the dielectric constant of the medium?
A. 5
B. 4
C. 3
D. 2

Answer 1

compare the given equation of electric field with the standard equation of electric field of the electromagnetic wave.

Formula used:
$c = \dfrac{\omega }{k}$

Here, $\omega$ is the angular frequency and k is the dielectric constant of the medium.
$c = \dfrac{1}{{\sqrt {\mu \varepsilon } }}$

Here, $\mu$ is the permeability of the free space and $\varepsilon$ is the dielectric constant of the air.

Complete step by step answer:
The electric field of the electromagnetic wave is given by the equation,$E = E_0 \sin \left( {\omega t - kx} \right)$ ...... (1)

Here, $E_0$ is the initial electric field, $\omega$ is the angular frequency, t is the time, k is the wave number, and x is the distance.

The given equation of the electric field is,$E = \left( {9 \times 10^8 \,NC^{ - 1} } \right)\sin \left[ {\left( {9 \times 10^8 \,rad\,s^{ - 1} } \right)t - \left( {6\,m^{ - 1} } \right)x} \right]$

Compare the above equation with the standard equation of electric field of the electromagnetic wave (1). We get,
$\omega = 9 \times 10^8 \,rad\,s^{ - 1}$and $k = 6\,m^{ - 1}$.

The speed of propagation of the wave is given by,
$c = \dfrac{\omega }{k}$
Substitute $9 \times 10^8 \,rad\,s^{ - 1}$ for $\omega$ and $6\,m^{ - 1}$ for $k$ in the above equation.
$c = \dfrac{{9 \times 10^8 \,rad\,s^{ - 1} }}{{6\,m^{ - 1} }}$

$c = 1.5 \times 10^8 \,ms^{ - 1}$

Also, the speed of electromagnetic wave in a dielectric medium of permittivity $\varepsilon$ is given by the equation,
$c = \dfrac{1}{{\sqrt {\mu \varepsilon } }}$

Here, $\mu$ is the permeability of the medium and $\varepsilon$ is the permittivity of the medium.

Therefore, we can write,
$\dfrac{1}{{\sqrt {\mu \varepsilon } }} = 1.5 \times 10^8 \,ms^{ - 1}$

Squaring the above equation, we get,
$\dfrac{1}{{\mu \varepsilon }} = 2.25 \times 10^{16} \,m^2 s^{ - 2}$

$\Rightarrow \varepsilon = \dfrac{1}{{2.25 \times 10^{16} \,m^2 s^{ - 2} \times \mu }}$

Substitute $4\pi \times 10^{ - 7} \,TmA^{ - 1}$ for $\mu$ in the above equation.
$\Rightarrow \varepsilon = \dfrac{1}{{2.25 \times 10^{16} \times 4\pi \times 10^{ - 7} }}$

$\varepsilon = \dfrac{1}{{28.26 \times 10^9 }}$

$\varepsilon = 0.035 \times 10^{ - 9} \,Fm^{ - 1}$

The dielectric constant of the medium is given by,
$k = \dfrac{\varepsilon }{{\varepsilon _0 }}$

Here, $\varepsilon _0$ is the permittivity of the free space and it has value $8.85 \times 10^{ - 12} \,Fm^{ - 1}$.

Substitute $0.035 \times 10^{ - 9} \,Fm^{ - 1}$ for $\varepsilon$ and $8.85 \times 10^{ - 12} \,Fm^{ - 1}$ for $\varepsilon _0$ in the above equation.$k = \dfrac{{0.035 \times 10^{ - 9} \,Fm^{ - 1} }}{{8.85 \times 10^{ - 12} \,Fm^{ - 1} }}$

$k = 3.95$

Therefore, the dielectric constant is nearly 4.

So, the correct answer is “Option B”.
Note:
Remember, the units of permittivity of the medium and permeability of the medium are in S.I. units. Also use $1\,T = kg\,s^{ - 2} A^{ - 1}$
wherever necessary if you don’t remember the units of the above-mentioned parameters.